This integral, in turn, can be estimated at \(\bar{\lambda }\rightarrow 0\)  making use of the expansion of the incomplete beta function [1]}:
\(\begin{split}\int _a^1dx x^{\alpha -1}(1-x)^{\beta -1}&=\frac{\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}-\sum _{n=0}^\infty \frac{(-1)^n\Gamma (\beta )a^{n+\alpha }}{n!\Gamma (\beta -n)(n+\alpha )},\\\int _a^1\frac{dx}{x}(1-x)^{\beta -1}&=-\gamma -\psi (\beta )-\ln a-\sum _{n=1}^\infty \frac{(-1)^n\Gamma (\beta )a^{n}}{n!\Gamma (\beta -n)n}.\end{split}\) 
